General Instructions: (i) The question paper comprises of two sections A and B. You are to attempt both the sections. (ii) All questions are compulsory. (iii) There is no overall choice. However internal choice has been provided in all the three question of five marks category. Only one option in such question is to be attempt. (iv) All questions of section A and all questions of section B are to be attempted separately. (v) Questions numbers 1 to 4 in section A are one mark questions. These are to be answered in one word or one sentence. (vi) Question numbers 5 to 13 are two mark questions, to be answered in about 30 words each. (vii) Question numbers 14 to 22 are three mark questions to be answered in about 50 words each. (viii) Question numbers 23 to 25 are five mark questions, to be answered in about 70 word each. (ix) Question numbers 26 to 41 in section B are multiple choice questions based on practical skills. Each question is a one mark question. You are to select one most appropriate response out of the four provided to you. SECTION - A Q 1: What are the various steps in a food chain called? Various levels in the food chain are called Trophic levels. Q 2: What is the important function of presence of ozone in earth s atmosphere? Ozone layer shield the earth s surface from the harmful effects of the UV radiation. Q 3: Write the electron dot structure of ethane molecule, C2H6. Page 1 of 21
The electron dot structure of ethane molecule, C2H6 is as follows: Q 4: What makes the earth s atmosphere a heterogeneous mixture? Earth s atmosphere is a mixture of different gases, it consists of 21% oxygen, 78% nitrogen, 0.93% argon and 0.03% CO2 along with other gases making up the remaining percentage. The composition varies at different places and times thus the Earth s atmosphere is heterogeneous mixture. Q 5: List any four characteristics of a good fuel. Characteristics of a good fuel are: i. It should possess low ignition temperature. ii. It should not leave residue after burning, i.e., it should burn completely. iii. It should burn without producing too many pollutants. iv. It should be easily available and accessible v. Good fuel must be economical. vi. It should be easy to store and transport. Q 6: What are non-renewable resources of energy? Give two examples of such resources. The energy sources that are consumed at a rate faster than the rate at which they are replenished are called Non-renewable resources. Page 2 of 21
E.g., Coal and petroleum. Q 7. (i) How do you calculate the possible valency of an element from the electronic configuration of its atoms? (ii) Calculate the valency of element X whose atomic number is 9. (i) The valency of an element is determined by the number of electrons present in the last shell. If, number of valence electrons is less than or equal to 4 then, valency = number of valence electrons. If, number of valence electrons greater than 4 then, valency = 8 number of valence electrons. (ii) The electronic configuration of element X with atomic number 9 is 2,7 Valency of X = 8 7 =1 Hence, the valency of element with atomic number 9 is 1. Q 8: On the basis of electronic configuration, how will you identify the first and the last element of a period? The element is said to be in the first element of a period if the number of valence electron is 1. If the number of valence electron is 8, the element is said to be the last element of a period. Q 9: State the two laws of reflection of light. Laws of reflection states that: i. The angle of incidence is equal to the angle of reflection. ii. The incident ray, the reflected ray and the normal to the mirror, all lie in the same plane. Q 10: The stars appear higher from horizon than they actually are. Explain why it is so. Page 3 of 21
There are different layers in the atmosphere. Due to variation of density in the atmospheric medium, refraction takes place. The atmosphere bends the starlight towards the normal. Hence, star appears slightly higher from the horizon. This apparent positioning is not stationary but keeps changing. Q 11: Explain why the planets do not twinkle but the stars twinkle. Planets are closer to the earth and do not appear to be point sources of light. They are considered to be extended sources of light. If, a collection of large number of point-sized sources of light is considered, then the total variation in the amount of light entering our eye from all the individual point sized sources will average out to zero, nullifying the twinkling effect. Stars on the other hand are point sized objects and twinkle by atmospheric refraction of light. Q 12: Write any two differences between binary fission and multiple fission in a tabular form as observed in cells of organisms. Difference between binary fission and multiple fission are: Binary Fission Multiple fission In binary fission, the single cell divides into halves In multiple fission, a single cell divides into that is into two. many daughter cells simultaneously. Binary fission is of two types; along any plane or Multiple fission occurs along only one plane. longitudinally. Example: Amoeba(along any Example: Plasmodium. plane), Leishmania (longitudinally) Q 13: Why is DNA copying an essential part of the process of reproduction? DNA copying is an important process that ensures that when cell multiplies, through the mechanisms of mitosis or meiosis, equal amount of DNA (genetic material) passes in to the new cell. DNA is copied so that the amount of DNA remains constant in the daughter cells. Q 14: Explain the terms: Page 4 of 21
(i) Speciation (ii) Natural selection (i) Speciation - Speciation is an evolutionary phenomenon by which new species are developed from existing ones. There are three models of speciation: 1. When the groups that evolve to be separate species are in different geographic locations and are isolated geographically from each other 2. When the groups that evolve to be separate species are geographic near and the individuals can move one species area to others. 3. When the groups that evolve to be separate species occur together in the same geographic area. (ii) Natural selection Natural selection is a key mechanism in evolution. It is a process that results in an increased survival and reproductive success of individuals that are well adjusted to the environment. The theory of natural selection was given by Charles Darwin. Q 15: Explain how equal genetic contribution of male and female parents is ensured in the progeny. All human beings have 23 pairs of chromosomes out of which 22 are autosomes and two chromosomes are sex chromosomes. Gametes are formed in the males and females by the process of meiosis and these gametes contain 23 chromosomes. The male gametes and female gametes are haploid and have 23 chromosomes. During fertilisation the male and the female gametes fuse to form zygote which is diploid and has 23 pairs of chromosomes. Thus both male and female show equal genetic contribution to the progeny Q 16: Out of HCl and CH3COOH, which one is a weak acid and why? Describe an activity to support your answer. HCl completely dissociates into ions in solution and CH3COOH gets partially dissociated into ions in solution. So, CH3COOH is a weak acid when compared to HCl. This can be proved by the following activity: 1. Two iron nails are fitted on a cork and are kept in a 100 ml beaker. 2. The nails are then connected to the two terminals of a 6-volt battery through a bulb and a switch. Page 5 of 21
3. Pour dilute HCl in the beaker and switch on the current. 4. Repeat the same procedure replacing HCl with CH3COOH. Observations: It will be observed that the bulb glows in the HCl solution and does not glow in the CH3COOH solution. Inference: HCl, being a strong acid dissociates into H + and Cl ions. These ions conduct electricity in the solution resulting in the glowing of the bulb. On the other hand, the CH3COOH is a weak acid. It does not dissociate into ions completely in solution. Therefore, it does not conduct electricity. Q 17: Two elements X and Y belong to group 1 and 2 respectively in the same period of periodic table. Compare them with respect to: (i) the number of valence electrons in their atoms (ii) their valencies (iii) metallic character (iv) the sizes of their atoms (v) the formulae of their oxides (vi) the formulae of their chlorides (i) The number of valence electrons in element X is 1. The number of valence electrons in element Y is 2. Page 6 of 21
(ii) The valency of an element is determined by the number of electrons in the outermost shell. Thus, the valency of element X is 1 and that of element Y is 2. (iii) Element X is more metallic than element Y. (iv) The size of atom X is more than that of atom Y. (v) The formula of oxide of element X is X2O and that of element Y is YO. (vi) The formula of chloride of element X is X Cl and that of element Y is YCl2. Q 18: Draw the ray diagram and also state the position, the relative size and the nature of image formed by a concave mirror when the object is placed at the centre of curvature of the mirror. The ray diagram when object is placed at the centre of curvature is given below: Position of the image: at the centre of curvature Size of the image: Enlarged Nature of the image: Real and inverted Q 19: (i) The refractive index of diamond is 2.42. What is the meaning of this statement? (ii) Name a liquid whose mass density is less than that of water but it is optically denser than water. (i) The refractive index of diamond is 2.42 implies that the ratio of sine of angle of incidence to the sine of angle of refraction is equal to 2.42. Page 7 of 21
(ii) Kerosene has refractive index of 1.44. Kerosene is optically denser than water, although its mass density is less than water. Q 20: What eye defect is hypermetropia? Describe with a ray diagram how this defect of vision can be corrected by using an appropriate lens. When a person cannot see the nearby objects distinctly but, can see the distant objects clearly then the person is suffering from hypermetropia. This defect can be corrected by using a convex lens of appropriate power. The below ray diagram illustrates the correction of this defect. Q 21: (a) List two sexually transmitted disease in each of the following cases: (i) Bacterial infections (ii) Viral infections (b) How may the spread of such diseases be prevented? Sexually transmitted disease in each of the following case is: (i) Bacterial infections: are Syphilis, Gonorrhoea. (ii) Viral infections: AIDS, warts. (b) Sexually transmitted disease can be prevented by using contraceptive measures like condom, avoiding sex with multiple partners or infected person. Page 8 of 21
Q 22: Explain Mendel s law of independent inheritance. Give one example. Mendel s law of independent assortment states that: When two pairs of traits are taken into consideration in a cross (dihybrid), one pair of character for a particular trait segregates independently of the other pair of character for different trait. For example: In a dihybrid cross between two plants having round yellow (RRYY) and wrinkled green seeds (rryy) two traits are considered colour of seed and shape of seed. In this four types of gametes (RY, Ry, ry, ry) are produced. Each of these segregate independent of each other, each having a frequency of 25% of the total gametes produced. Thus the characters for colour of the seed segregate independently from the characters of shape of the seed. Q 23: (a) If the image formed by a lens is diminished in size and erect, for all positions of the object, what type of lens is it? (b) Name the point on the lens through which a ray of light passes undeviated. Page 9 of 21
(c) An object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find (i) the position (ii) the magnification and (iii) the nature of the image formed. OR (a) One-half of a convex lens is covered with a black paper. Will such a lens produce an image of the complete object? Support your answer with a ray diagram. (b) An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm. (i) Draw the ray diagram and (ii) Calculate the position and size of the image formed. (iii) What is the nature of the image? (a) If the image formed by a lens is diminished in size and erect, for all positions of the object, then the lens is a concave lens. (b) The point on the lens through which a ray of light passes undeviated is known as Pole. (c) Given, Object distance, u = -30 cm Focal length, f =20 cm i) Now, using the len s formula, 1 f = 1 v 1 u 1 v = 1 f + 1 u v = u f u+f v = ( 30) 20 20 30 v = 60 cm Page 10 of 21
The image is formed at a distance of 60 cm on the other side of the optical centre (ii) Magnification, m = m = v u = 60 30 = 2 (iii) Image formed is inverted. OR (a) Image is formed by a large number of rays from the object. If one part of the lens is blackened, image will be formed. But, intensity of the image will be reduced. (b) i) Object distance, u = -25 cm Focal length, f = 10 cm Height of the image, h = 5 cm ii) Now, using the lens formula, 1 f = 1 v 1 u We have, 1 v = 1 f + 1 u v = u f u+f Page 11 of 21
v = 25 10 10 25 Now, magnification = 16.67 cm (iii) Negative sign indicates that the image is real & inverted. Q 24: (a) Draw a diagram of the longitudinal section of a flower and label on it sepal, petal, ovary and stigma. (b) Write the names of male and female reproductive parts of a flower. OR (a) What is fragmentation in organisms? Name a multicellular organism which reproduces by this method. (b) What is regeneration in organism? Describe regeneration in Planaria with the help of a suitable diagram. (a) Diagram of longitudinal section of flower is shown below: (b) Male reproductive part of the flower is the stamen which consists of filaments and anthers. The female reproductive part of the flower is the pistil that consists of stigma, style and ovary. OR Page 12 of 21
(a) Fragmentation is a mode of asexual reproduction. Fragmentation is the process in certain multi-cellular organisms with relatively simple body organisation, the body of the organism simply breaks up into smaller pieces upon maturation and these pieces or fragments grow into new individuals. Example: Spirogyra. (b) Regeneration: Regeneration involves the capacity of an organism to give rise to an entire individual from a cut portion. This type of regeneration occurs in Planaria. When Planaria gets cut unintentionally, the cut fragments give rise to a new individual. Regeneration is carried out by specialised cells. These cells proliferate and make large numbers of cells. From this mass of cells, different cells undergo changes to become various cell types and tissues. The figure shows regeneration in planaria: Q 25: (a) In a tabular form, differentiate between ethanol and ethanoic acid under the following heads: (i) Physical state (ii) Taste (iii) NaHCO3 test (iv) Ester test (b) Write a chemical reaction to show the dehydration of ethanol. OR (a) What is soap? Why are soaps not suitable for washing clothes when the water is hard? (b) Explain the action of soap in removing an oily spot from a piece of cloth. Page 13 of 21
(a) Property Ethanol Ethanoic acid (i) Physical state Ethanol is a colorless liquid with Ethanoic acid is colorless, pungent pleasant odor smelling liquid (ii) Taste Ethanol is bitter to taste Ethanoic acid is sour to taste (iii) NaHCO3 test Ethanol does not react with When ethanoic acid reacts with sodium sodium bicarbonate NaHCO3 with the evolution of carbon dioxide gas. (iv) Ester test Ethanol on reaction with Ethanoic acid on reaction with alcohols in ethanoic acid in the presence of the presence of conc. Sulphuric acid to acid forms ester. form ester. (b) Ethanol undergoes dehydration to form ethane. OR 2 CH 3 CH 2 OH Conc.H 2 SO 4 2CH 2 = CH 2 + 2H 2 O (a) Soap is sodium or potassium salt of higher fatty acids such as oleic acid (C17H33COOH), stearic acid (C17H35COOH), palimitic acid (C15H31COOH), etc. A soap is a sodium or potassium salt of long chain fatty acids. Hard water contains salts of calcium and magnesium. On adding soap to water, calcium and magnesium ions present in water displace sodium or potassium ions from the soap molecules forming an insoluble substance called scum. Scum results in wastage of soap. (b) Cleansing action of soaps: The oily spot present on clothes is organic in nature and insoluble in water. Therefore, it cannot be removed by only washing with water. When soap is dissolved in water, its hydrophobic ends attach themselves to the oily spot and remove it from the cloth. Then, the molecules of soap arrange themselves in the form of micelle and trap the dirt at the centre of the cluster. These micelles remain suspended in the water. Hence, the oily spots are easily rinsed away by water. Page 14 of 21
Section - B Q 26: A student was given two permanent slides, one of binary fission in amoeba and other of budding in yeast. He was asked to identify any one difference in the nucleus of the two. One such difference, he identified correctly was (1) Presence of one nucleus in amoeba, two in yeast cell and one in bud. (2) Presence of two nuclei in centrally constricted amoeba, one in yeast cell and one in its bud. (3) Presence of two distant nuclei in amoeba, one in yeast cell and two in bud. (4) Presence of a single nucleus each in amoeba, yeast cell and its attached bud. The correct option is (2). Presence of two nuclei in centrally constricted amoeba, one in yeast cell and one in its bud. Q 27: Binary fission is observed in which one of the following figures? Page 15 of 21
(1) A (2) B (3) C (4) D (3) C Binary fission in Amoeba. Q 28: To determine the percentage of water absorbed by raisins, raisins are soaked in water for: (1) 30 seconds (2) 10 minutes (3) 2 to 3 hours (4) 24 hours The correct option is (3) 2 to 3 hours Q 29: Raisins are wiped off gently before final weighing with help of (1) a filter paper (2) a cotton piece (3) a cloth piece (4) a polythene piece The correct option is (1) a filter paper Q 30: The step(s) necessary for determining the percentage of water absorbed by raisins is/are: (1) Raisins should be completely immersed in water Page 16 of 21
(2) Raisins should be soaked in water for sufficient time (3) Gently wipe dry the soaked raisins (4) All of the above steps. The correct option is (4) All of the above steps. Q 31: Rahim recorded the following sets of observations while tracing the path of a ray of light passing through a rectangular glass slab for different angles of incidence. S. No. Angle of incidence Angle of refraction Angle of emergence I 45 o 41 o 45 o II 40 o 38 o 38 o III 45 o 41 o 40 o IV 41 o 45 o 41 o The correct observations are recorded at a serial number: (1) I (2) II (3) III (4) IV The correct set of observation is (1). Q 32: Four students A, B, C and D traced the paths of incident ray and the emergent ray by fixing pins P and Q for incident ray and pins R and S for emergent ray for a ray of light passing through a glass slab. Page 17 of 21
The correct emergent ray was traced by the student: (1) A (2) B (3) C (4) D The correct emergent ray is traced by student B. So the correct option is (2). Q 33: Mohan obtained a sharp inverted image of a distant tree on the screen placed behind the lens. He then moved the screen and tried to look through the lens in the direction of the object. He would see: (1) a blurred image on the wall of the laboratory. (2) an erect image of the tree on the lens. (3) no image as the screen has been removed Page 18 of 21
(4) an inverted image of the tree at the focus of the lens. The correct option is (1) a blurred image on the wall of the laboratory. Q 34: For finding the focal length of a convex lens by obtaining the image of a distant object, one should use as the object. (1) A well-lit distant tree (2) Window grill in the class room (3) Any distant tree (4) A lighted candle kept at the other end of the table. The correct option is (1) a well-lit distant tree. Q 35: To find the focal length of a concave mirror Rahul focuses a distant object with this mirror. The chosen object should be (1) A tree (2) A building (3) A window (4) The sun The correct option is (4) the sun. Q 36: The color of an aqueous solution of zinc sulphate as observed in the laboratory is: (1) Green (2) Yellow (3) Blue (4) Colorless Page 19 of 21
The correct option is (4) Colorless. Q 37: To show that zinc is a more active metal than copper, the correct procedure is to: (1) Add dilute nitric acid on strips of both the metals. (2) Observe transmission of heat through strips of zinc and copper. (3) Prepare solution of zinc sulphate and hang strip of copper into it. (4) Prepare solution of copper sulphate and hang strip of zinc into it. The correct option is (4) Prepare solution of copper sulphate and hang strip of zinc into it. Q 38: Acetic acid smells like: (1) a banana (2) vinegar (3) an orange (4) a lemon The correct option is (2) vinegar. Q 39: Acetic acid solution turns: (1) blue litmus red (2) red litmus blue (3) blue litmus colourless (4) red litmus colourless The correct option is (1) blue litmus red. Page 20 of 21
Q 40: On adding NaHCO3 to acetic acid, a gas is evolved which turns lime water milky due to the formation of: (1) Calcium bicarbonate (2) Calcium hydroxide (3) Calcium carbonate (4) Calcium acetate The correct option is (3) Calcium carbonate. Q 41. A yeast cell in which budding occurs was seen to have: (1) one bud cell (2) two bud cell (3) three bud cell (4) a chain of bud cells The correct option is (1) one bud cell. Page 21 of 21